Q. Let $9 = \mathrm{x_1} < \mathrm{x_2} < ... < \mathrm{x_7}$ be in an A.P. with common difference $d$. If the standard deviation of $x_1, x_2 ..., x_7$ is 4 and the mean is $\overline{x}$, then $\overline{x}+x_6$ is equal to :
A) $18\left(1+\frac{1}{\sqrt{3}}\right)$
B) $34$
C) $2\left(9+\frac{8}{\sqrt{7}}\right)$
D) $25$
Explanation
The correct answer is Option B: $34$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.